Showing $x^4+x^2+x+1$ irreducible over $\mathbb{Z_3}$

Question

$x^4+x^2+x+1$ irreducible over $\mathbb{Z_3}$. So since there are no roots there are no linear factors. From here do you just try to factor it as a product of 2 quadratics and show it that this leads to a contradiction? Is that the only way? Furthermore can we assume these quadratics are monic? why? Thanks yall!

Answer 1

Yes, solving a system of equations is the only way to show that the polynomial is irreducible. You can, however, assume that both factors are monic:

If $f(x)=x^4+x^2+x+1$ factors as $f(x)=p(x)q(x)$, then either both $p$ and $q$ are monic, or both have lead coefficient $2$. In the latter case, you have $f(x)=2p(x)2q(x)$ and both $2p(x)$ and $2q(x)$ are monic.